Answer :

To find out how many milliliters of a 1.55 M NaBr solution are needed to contain 75.0 grams of NaBr, we can follow these steps:

1. Calculate the Moles of NaBr:

First, we need to determine how many moles of NaBr are present in 75.0 grams. To do this, we use the molecular weight of NaBr, which is 102.89 g/mol.

[tex]\[
\text{Moles of NaBr} = \frac{\text{Mass of NaBr (g)}}{\text{Molecular Weight of NaBr (g/mol)}} = \frac{75.0 \, \text{g}}{102.89 \, \text{g/mol}} \approx 0.729 \, \text{mol}
\][/tex]

2. Calculate the Volume in Liters:

Next, we use the definition of molarity, which is moles of solute per liter of solution (M = mol/L), to find the volume of the solution in liters. We know the molarity of the solution is 1.55 M.

[tex]\[
\text{Volume (L)} = \frac{\text{Moles of NaBr}}{\text{Molarity}} = \frac{0.729 \, \text{mol}}{1.55 \, \text{M}} \approx 0.470 \, \text{L}
\][/tex]

3. Convert Volume from Liters to Milliliters:

Since there are 1000 milliliters in a liter, we convert the volume from liters to milliliters.

[tex]\[
\text{Volume (mL)} = 0.470 \, \text{L} \times 1000 \, \text{mL/L} = 470 \, \text{mL}
\][/tex]

Therefore, 470 milliliters of a 1.55 M NaBr solution contain 75.0 grams of NaBr.

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Rewritten by : Barada